CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
The function f(x) = x³ − 12x + 5 has a local minimum at x equal to:
The local minimum occurs at x = 2. Setting f'(x) = 3x² − 12 to zero gives x = ±2, and the second derivative 6x is positive only at x = 2, so that point is a minimum, while x = −2 is a maximum.
- Ax = −2
- Bx = 0
- Cx = 2Correct
- Dx = 4
Explanation
f'(x) = 3x² − 12 = 0 gives x = ±2. f''(x) = 6x, which is 12 > 0 at x = 2 (minimum) and −12 < 0 at x = −2 (maximum). So the local minimum is at x = 2; x = −2 is the local maximum.
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